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Around 1869∼1872, Charles M´eray ([41]), Georg Cantor ([10]) and Eduard Heine ([28]) independently introduced basically the same additive operation ⊞, multiplicatived operation ⊠ and total order on CR, getting yet another number system (CR, ⊞, ⊠, ). This approach has the advantage of providing a standard way for completingb an abstract metricb space. b d d b b b Both systems are isomorphic in the sense that one can find a bijection ω : DR → CR such that for any two elements x,y ∈ DR, ω(x⊞y)= ω(x)⊞ ω(y), ω(x⊠y)= ω(x)⊠ ω(y), x y ⇔ ω(x) ω(y). Detailed studies of a dozen or so properties of ⊞, ⊠, lead tod the basic concept of “complete ordered field” and an algebraic-axiomaticb approach to theb real number system.b Later on, when to verify a new system is isomorphic to those of Dedekind and M´eray-Cantor-Heine, it suffices to prove that it is a complete ordered field. We should also note several severe criticisms of the algebraic-axiomatic approach: “The necessary axioms should come as a byproduct of the construction process and not be predetermined.” ([38]) “. . . the algebraic-axiomatic definition of a real number is simply appalling and abhorrent . . . to define a real number via a cold and boring list of a dozen or so axioms for a complete ordered field is like replacing life by death or reading an obituary column.” ([3]) ∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼ “Few mathematical structures have undergone as many revisions or have been presented in as many guises as the real numbers. Every generation reexamines the reals in the light of its values and mathematical objectives.” ([21]) Next we introduce the real numbers through a rather old geometric approach. Given a point x in an “axis”, the decimal representation of it is obtained step by step as follows.
• Step 0: Partition this “axis” into countably many disjoint unions z∈Z[z, z + 1), then find a unique integer x0 ∈ Z such that x ∈ [x0,x0 + 1). 9 i S i+1 • Step 1: Partition [x0,x0 + 1) into ten disjoint unions i=0[x0 + 10 ,x0 + 10 ), then find a unique element x ∈ Z such that x ∈ [x + x1 ,x + x1+1 ), here and 1 10 0S 10 0 10 afterwards we denote {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} by Z10 for simplicity. x1 x1+1 9 x1 • Step 2: Partition [x0 + 10 ,x0 + 10 ) into ten disjoint unions j=0[x0 + 10 + j ,x + x1 + j+1 ), then find a unique element x ∈Z such that x ∈ [x + x1 + 102 0 10 102 2 10 S 0 10 x2 x1 x2+1 102 ,x0 + 10 + 102 ). . • Step .:.
Then we say x has decimal representation x0.x1x2x3 , and call xk the k-th digit of x. A thorough consideration leads to a natural question: can any point of this “axis” have decimal representation 0.999999 ? and if not, why can we expel its existence? This is not a silly question at all. To correctly answer it, we should first know what an “axis” it is, or what on earth a “real number” it is! Although constructing real numbers via decimals has been known since Simon Stevin ([1, 2]) in the 16-th century, developed also by Karl Weierstrass ([13]), Otto Stolz ([50, 51]) in the 19-th century, and many other modern mathematicians (see e.g. [3, 5, 8, 9, 14, 21, 24, 26, 30, 34, 37, 38, 40, 44, 54]), almost all popular Mathematical Analysis books didn’t choose this approach. The decimal construction hasn’t got the attention it should deserve: ANEWAPPROACHTOTHEREALNUMBERS 3
“Perhaps one of the greatest achievements of the human intellect throughout the entire history of the human civilization is the introduction of the decimal notation for the purpose of recording the measurements of various magnitudes. For that purpose the decimal notation is most practical, most simple, and in addition, it reflects most outstandingly the profound subtleties of the human analytic mind. In fact, decimal notation reflects so much of the Arithmetic and so much of the Mathematical Analysis . . .” ([3]) To the author’s opinion, at least one of the reasons behind this phenomenon is most of the authors only gave outlines or sketches. We note an interesting phenomenon happened in popular Mathematical Analysis books: many authors (see e.g. [4, 7, 11, 22, 31, 35, 36, 42, 43, 45, 47, 48, 49, 52, 55]) first chose one of the other three approaches discussed before, then proved that every real number has a suitable decimal representation. But if without using the algebraic intuition that the real number system is nothing but a complete ordered field which is not so easy to grasp for beginners, we could find basically no literature reversing this kind of discussion. No matter how fundamental sets and sequences are in mathematics, conflicting with our primary, secondary and high school education that a number is a string of decimals, skepticism over explicit construction of the real numbers by either Dedekind cuts or Cauchy sequences never ends: “The degeometrization of the real numbers was not carried out without skepticism. In his opus Mathematical Thought from Ancient to Modern Times, mathematics historian Morris Klein quotes Hermann Hankel who wrote in 1867: Every attempt to treat the irrational numbers formally and without the concept of [geometric] magnitude must lead to the most abstruse ad troublesome artificialities, which, even if they can be carried through complete rigor, as we have every right to doubt, do not have a right scientific value.” ([24]) “The definition of a real number as a Dedekind cut of rational numbers, as well as a Cauchy sequence of rational numbers, is cumbersome, impractical, and . . ., inconsequential for the development of the Calculus or the Real Analysis.” ([3]) Based on the fundamental concept of “order” and its derived operations, in this paper we will provide a complete approach to the real numbers via decimals, and some of our ideas are new to the existing literatures. Also in this new setting, construction of the real numbers by Dedekind cuts, Cauchy sequences of rational numbers, and the algebraic characterization of the real number system by the concept of complete ordered field can be well explained. The general strategy of our approach is as follows. The starting point is to choose
R x0.x1x2x3 x0 ∈ Z,xk ∈Z10, k ∈ N . as our ambient space, which was already discussed before. Once adopted this decimal ∞ xi notation, we suggest you imagine it in mind as a series k=0 10i . There are mainly two reasons why we choose this notation, one is instead of discussing “subtraction”, we shall focus on “additive inverse” which will be introduced in anP elegant manner, the other comes from the next paragraph. N Since as sets R is basically the same as Z×Z10, we can introduce a lexicographical order on R, then prove the least upper bound property and the greatest lower bound property for (R, ) from the same properties for (Z, ≤) as soon as possible. As experienced readers should know, this would mean that (R, ) is “complete”. So in this complete setting, we can derive five basic operations such as the supremum operation sup( ), the infimum operation inf( ), the upper limit operation LIMIT( ), the lower limit operation LIMIT( ) and the limit operation LIMIT( ). 4 LIANGPAN LI
It is very natural to define
x ⊕ y LIMIT {[x]k + [y]k}k∈N Pk x 10k−i i=0 i Q for any two elements x,y ∈ R, where [x]k x0.x1x2 xk = 10k ∈ is the truncation of x up to the k-th digit, so is [y]k. As for the definition of [x]k + [y]k, even a primary school student may know how to do it, that is, for example, (−15).3456 +(−18).6789 (−32).0245
To define a multiplicative operation in a succinct way we need some preparation. First we introduce a signal map sign : R → {0, 1} by 0 if x 0.000000 , sign(x) 1 if x (−1).999999 . This map partitions R into two parts, one is sign−1(0), understood as the positive part of R, the other is sign−1(1), understood as the negative part of R. Both parts are closed connected through an “additive inverse” map
Ψ(x0.x1x2x3 ) (−1 − x0).(9 − x1)(9 − x2)(9 − x3) , which turns a positive element into a negative one, and vice versa. The absolute value of x, denoted by x , is defined to be the maximum over x and Ψ(x). Because of the wonderful formula x =Ψ(sign(x))( x ), the author likes to call them three golden flowers. Now for any two elements x,y ∈ R, we define their multiplication by sign(x)+sign(y) x ⊗ y Ψ LIMIT {[ x ]k [ y ]k}k∈N . In primary school we have already learnt how to define [ x ]k [ y ]k. We remark that motivated by the above definitions of addition and multiplication on R, similar operations will be introduced on DR in a highly consistent way. At this stage, we have introduced a rough system (R, , ⊕, ⊗) without any pain. But unfortunately, several well-known properties generally a standard addition and a standard multiplication should have don’t hold for ⊕ and ⊗. To overcome these difficulties, we will introduce an equivalence relation ∼ identifying 0.999999 with 1.000000 , and the same like. No matter adopting decimal, binary or hexadecimal notation, no matter introducing such relations earlier or later, we cannot avoid doing it. With these preparations, we then verify the commutative, associate and distribute laws, the existences of additive (multiplicative) unit and inverse, and so on. Almost all the verification work depend only on a pleasant Lemma 3.8. Finally we define the set R of real numbers to be the set of equivalent classes R/ ∼ with derived operations ⊕, ⊗, from ⊕, ⊗, respectively, thus yields our desired number system (R, ⊕, ⊗, ). b Now in the new setting (R, ) with derived operations such as sup( ), inf(b b) for subsets, and LIMIT( ) and LIMIT( ) for sequences of R, we can furtherb b explainb the construction of the real numbers by Dedekind cuts and Cauchy sequences of rational numbers. Given a Dedekind cut (A|B), we obviously have sup A inf B. It would be very nice if sup A ∼ inf B, thus one can derive a map τ from DR to R by sending (A|B) to [sup A] = [inf B]. Later on we shall prove that this is indeed the case. Given a Cauchy sequence of rational ANEWAPPROACHTOTHEREALNUMBERS 5
(n) (n) (n) numbers {x }n∈N, we obviously have LIMIT({x }n∈N) LIMIT({x }n∈N). It would (n) (n) be great if LIMIT({x }n∈N) ∼ LIMIT({x }n∈N), thus one can derive a map κ from (n) (n) (n) CR to R by sending {x }n∈N to [LIMIT({x }n∈N)] = [LIMIT({x }n∈N)]. Later on we shall prove that this is also indeed the case. Motivated by these observations, we will continue to prove that our number system is isomorphic to those of Dedekind and M´eray-Cantor-Heine. We can explain Cauchy sequences of rational numbers in another way. Given a Cauchy (n) sequence of rational numbers {x }n∈N, if it could represent a “real number”, then we have no doubt that any subsequence of it, say for example a monotonically increasing or a monotonically decreasing subsequence, should represent the same “real number”. Then we put such a monotone subsequence of rational numbers in (R, ) whose existence is taken for granted at this time, from either the least upper bound property or the greatest lower bound property in the new setting, the interested readers definitely know which decimal representation should be understood as the “real number” the original sequence represents. Simply speaking, it would be great if we can discuss Dedekind cuts or Cauchy sequences of rational numbers in a constructive, complete setting. Still in the setting (R, ), we will give the traditional characterization of irrational numbers, which makes our approach matching what we have learnt in high school. We also explain how can we come to the basic concept of complete ordered field. ∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼∼ To develop this approach, the author owed a lot to Loo-Keng Hua’s masterpiece [30]. He also thanks Rong Ma for helpful discussions. This work was partially supported by the Natural Science Foundation of China (Grant Number 11001174). Some notations used throughout this paper: • N is the set of natural numbers • Z is the set of integers • Denote {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} by Z10 • ⌊ ⌋ is the floor function, ⌈ ⌉ is the ceil function • For any nonempty bounded above (below) subset E of Z, denote by max E (inf E) the unique least upper (greatest lower) bound for E • For any binary operation ⊛ on a set G, denote by A⊛B the set {a⊛b| a ∈ A, b ∈ B}, where A, B are nonempty subsets of G • Rational number system: (Q, +, ×, ≤) • Decimal system: (R, ⊕, ⊗, ) • Dedekind cut system: (DR, ⊞, ⊠, ) • Cauchy sequence system: (CR, ⊞, ⊠, ) • Real number system: (R, ⊕, ⊗, ) d b b b 2. Least upper bound andb greatestb b lower bound properties for (R, )
2.1. Ambient space, lexicographical order and its derived operations. Definition 2.1. As in Section 1 we define the ambient space of this paper to be
R x0.x1x2x3 x0 ∈ Z,xk ∈Z10, k ∈ N . Pk x 10k−i N i=0 i Q For any x = x0.x1x2x3 ∈ R and k ∈ , let [x]k x0.x1 xk = 10k ∈ be the truncation of x up to the k-th digit. 6 LIANGPAN LI
q Remark 2.2. For any α ∈ Q, there exist two integers p ∈ N and q ∈ Z such that α = p . Via the long division algorithm α has a decimal representation α0.α1α2α3 , where q 10β α ⌊ ⌋, β q − α p, α ⌊ k ⌋, β 10β − α p. 0 p 0 0 k+1 p k+1 k k+1 Obviously, this representation is independent of the choices of p and q, so from now on we can always view Q as a subset of R. For example, 5 −40 = 2.5000000 , = (−14).666666 . 2 3 Definition 2.3. Let be the lexicographical order on R, that is, x y if and only if ∀k ∈ N, [x]k ≤ [y]k.
As usual, we may write y x when x y, and x ≺ y when x y, x = y. Obviously, as elements of Q, x ≤ y if and only if x y, which is self-evident since we can write x,y by a common denumerator, then use the long division algorithm. Definition 2.4. A nonempty subset W of R is called bounded above (below) if there exists an M ∈ R such that ∀w ∈ W , w M (w M). A nonempty subset of R is called (n) bounded if it is bounded both above and below. A sequence {x }n∈N of R is called bounded above (below), bounded if the set {x(n)| n ∈ N} is of the corresponding property. Theorem 2.5. Every nonempty bounded above subset of (R, ) has a least upper bound, every nonempty bounded below subset of (R, ) has a greatest lower bound.
Proof. This theorem follows from the least upper bound property and the greatest lower bound property for (Z, ≤). We shall only prove the first part of this theorem, and leave the second one to the interested readers. Let W be a nonempty bounded above subset of R. Denote M0 max{x0| x0.x1x2x3 ∈ W },
Mk max{xk| x0.x1x2x3 ∈ W with xi = Mi, i = 0, 1,...,k − 1} (∀k ∈ N).
Obviously, M M0.M1M2M3 is an upper bound for W . Let M be an arbitrary upper bound for W . ∀k ∈ N, by the definition of Mk one can find an x = x0.x1x2x3 ∈ W such that xi = Mi for i = 0, 1,...,k. Thus [M]k ≥ [x]k = x0.x1 xk =fM0.M1 Mk = [M]k, which means M M. This proves M is the (unique) least upper bound for W . f f Definition 2.6. The least upper bound, also known as the supremum, for a nonempty bounded above subset W of R is denoted by sup W , while the greatest lower bound, known as the infimum, for a nonempty bounded below subset W of R is denoted by inf W . (n) Definition 2.7. Generally given a sequence {x } N, we should pay attention to its n∈f f asymptotic behavior. Obviously for any n ∈ N, sup{x(k)| k ≥ n} can be understood as (n) an “upper bound” for the sequence {x }n∈N if we don’t care about several of its initial terms. Thus if we want to obtain an asymptotic “upper bound” that is as least as possible, then we are naturally led to the concept of upper limit of a bounded sequence, that is, (n) (k) LIMIT({x }n∈N) inf sup{x | k ≥ n} n ∈ N . (n) Similarly, we define the lower limit of a bounded sequence {x }n∈N to be
(n) (k) LIMIT({x }n∈N) sup inf{x | k ≥ n} n ∈ N .
ANEWAPPROACHTOTHEREALNUMBERS 7
(n) (n) When it happens that LIMIT({x }n∈N) = LIMIT({x }n∈N)= L, we shall simply write (n) (n) LIMIT({x }n∈N) for their common value L, and say the sequence {x }n∈N has limit L. Remark 2.8. It is no hard to prove that any bounded above monotonically increasing sequence, that is, x(1) x(2) x(3) x(4) M, or any bounded below monotoni- cally decreasing sequence, that is, x(1) x(2) x(3) x(4) M, has a limit. For example, suppose x(1) x(2) x(3) x(4) M. Then (n) (k) LIMIT({x }n∈N) = inf sup{x | k ≥ n} n ∈ N (k) = inf sup{x | k ∈ N} n ∈ N
(k) = sup{x | k ∈ N}
= sup inf{x(n)| n ≥ k} k ∈ N (n) = LIMIT ({x }n∈N).
We state below some elementary properties of sup( ), inf( ), LIMIT( ), LIMIT( ) and LIMIT( ) in a lemma, and leave the proofs to the interested readers. Lemma 2.9. (1) inf W sup W , (2) W1 ⊂ W2 ⇒ sup W1 sup W2, (3) W1 ⊂ W2 ⇒ inf W1 inf W2, (n) (n) (4) LIMIT({x }n∈N) LIMIT({x }n∈N), (n) (n) (n) (n) (5) x y ⇒ LIMIT({x }n∈N) LIMIT({y }n∈N), (n) (n) (n) (n) (6) x y ⇒ LIMIT({x }n∈N) LIMIT({y }n∈N), (ni) (n) (7) LIMIT {x }i∈N LIMIT {x }n∈N , (8) LIMIT {x(ni)} LIMIT {x(n)} , i∈N n∈N (9) LIMIT {x(n)} = L ⇒ LIMIT {x(ni)} = L, n∈N i∈N (10) LIMIT {[x] } = x. k k∈N 2.2. An equivalence relation. Definition 2.10. Two elements x,y of R are said to have no gap, denoted by x ∼ y, if it does not exist an element z ∈ R\{x,y} lying exactly between x and y.
Definition 2.11. Let R9, QF be respectively the sets of all decimal representations ending in an infinite string of nines, and zeros.
Lemma 2.12. x ≺ y, x ∼ y ⇒ x ∈ R9,y ∈ QF .
Proof. Suppose x = x0.x1x2x3 ≺ y = y0.y1y2y3 . Let k be the minimal non-negative integer such that xk x x0.x1 xk999999 ≺ y0.y1 yk000000 y. Since x ∼ y, we must have x = x0.x1 xk999999 , else we get a contradiction x ≺ x0.x1 xk999999 ≺ y. Similarly, y = y0.y1 yk000000 . This finishes the proof. Remark 2.13. As a corollary of Lemma 2.12, it is easy to prove that ∼ is an equivalence relation on R, and every equivalent class has at most two elements of R. As we know in real life, one can find a medium of any two different points in a straight line, but at this time it is hard for us to define the medium between two points having no gap. So to express a straight line from R, it is absolutely necessary to module something from R, which we shall discuss in detail at a very late stage of this paper. 8 LIANGPAN LI Next we prepare a useful characterization lemma. An enhanced Lemma 3.8 will be given in the next section. Lemma 2.14. N 1 x ∼ y ⇔∀k ∈ , |[x]k − [y]k|≤ 10k . Proof. The necessary part follows immediately from Lemma 2.12, so we need only prove N 1 the sufficient one, and suppose ∀k ∈ , |[x]k − [y]k|≤ 10k . Without loss of generality we N 1 may assume that x y. Thus ∀k ∈ , [x]k ≤ [y]k ≤ [x]k + 10k . Case 1: Suppose x ∈ R9. There exists an m ∈ N such that ∀k >m,xk = 9. Obviously, 1 x ∼ x0.x1x2 xm + 10m . Note ∀k>m, 1 1 [x] ≤ [y] ≤ [x] + = x .x x x + . k k k 10k 0 1 2 m 10m 1 Letting k →∞ gives x y x0.x1x2 xm + 10m , here we have used the fifth and tenth parts of Lemma 2.9. This naturally implies x ∼ y. Case 2: Suppose x ∈ R9. There exists a sequence of natural numbers m1 < m2 < m3 < such that ∀i ∈ N, xmi < 9. Noting 1 [x]m ≤ [y]m ≤ [x]m + i i i 10mi and 1 [x]m = [x]m + = [x]m −1, i mi−1 i 10mi mi−1 i we must have [y]m −1 = [y]m = [x]m −1. By the ninth and tenth parts of Lemma i i mi−1 i 2.9, we have x = y. This concludes the whole proof of the lemma. 3. Additive operations 3.1. Additive operations. Definition 3.1 (Addition). For any two elements x,y ∈ R, let x ⊕ y LIMIT {[x]k + [y]k}k∈N . This is well-defined since the sequence {[x] k + [y]k}k∈N is monotonically increasing with an upper bound x0 + y0 + 2, where x = x0.x1x2x3 , y = y0.y1y2y3 . Example 3.2. For any x ∈ R, x ⊕ 0.000000 = x. This means (R, ⊕) has a unit. Remark 3.3. Given two elements x,y ∈ QF ⊂ Q, it is easy to verify that x ⊕ y = x + y. Therefore from now on we can abuse the uses of ⊕ and + if the summands lie in QF . Theorem 3.4. For any two elements x,y ∈ R, we have x ⊕ y = y ⊕ x. Proof. x ⊕ y = LIMIT {[x]k + [y]k}k∈N = LIMIT {[y]k + [x]k}k∈N = y ⊕ x. Theorem 3.5. Given x, y, z, w ∈ R with x z and y w, we have x ⊕ y z ⊕ w. Proof. ∀k ∈ N we have [x]k ≤ [z]k and [y]k ≤ [w]k, which yields [x]k + [y]k ≤ [z]k + [w]k. By Lemma 2.9, LIMIT {[x]k + [y]k}k∈N LIMIT {[z]k + [w]k}k∈N . Lemma 3.6. N 1 For any element x ∈ R and k ∈ , we have [x]k x [x]k + 10k . ANEWAPPROACHTOTHEREALNUMBERS 9 Proof. The first inequality is evident and we need only to prove the second one. For any 1 1 natural numbers n ≥ k, [x]n ≤ [x]k + 10k . Letting n →∞ yields x [x]k + 10k . Lemma 3.7. For any elements x,y ∈ R and k ∈ N, we have 2 [x] + [y] ≤ [x ⊕ y] x ⊕ y [x] + [y] + . k k k k k 10k Proof. To prove the first inequality, we need only note [x]k + [y]k x ⊕ y ⇒ [x]k + [y]k = [[x]k + [y]k]k ≤ [x ⊕ y]k. For any natural numbers n ≥ k, 1 1 2 [x] + [y] ≤ ([x] + ) + ([y] + ) = [x] + [y] + . n n k 10k k 10k k k 10k 2 Letting n → ∞ yields x ⊕ y [x]k + [y]k + 10k . This proves the third inequality. The second one follows from the previous lemma, so we finishes the whole proof. Lemma 3.8. N N M If there exists an M ∈ such that ∀k ∈ , |[x]k − [y]k|≤ 10k , then x ∼ y. N 1 M+1 M+1 N Proof. For any k ∈ , y [y]k + 10k ≤ [x]k + 10k x ⊕ 10k . For any m ∈ , we can M+1 1 1 find a sufficiently large k such that 10k ≤ 10m . Consequently, y x ⊕ 10m , which yields 1 1 1 [y]m ≤ [x ⊕ 10m ]m = [x]m + 10m . By symmetry we can also have [x]m ≤ [y]m + 10m . Thus 1 |[x]m − [y]m|≤ 10m , and this finishes the proof simply by applying Lemma 2.14. Theorem 3.9. Given x, y, z, w ∈ R with x ∼ z and y ∼ w, we have x ⊕ y ∼ z ⊕ w. Proof. For any k ∈ N, 2 4 4 [x ⊕ y] ≤ [x] + [y] + ≤ [z] + [w] + ≤ [z ⊕ w] + , k k k 10k k k 10k k 10k here we have used Lemma 2.14 for x ∼ z and y ∼ w. By symmetry we can also have 4 [z ⊕ w]k ≤ [x ⊕ y]k + 10k . Finally by Lemma 3.8, we are done. Theorem 3.10. For any three elements x,y,z ∈ R, we have (x ⊕ y) ⊕ z ∼ x ⊕ (y ⊕ z). Proof. For any k ∈ N, [x]k + [y]k + [z]k ≤ [x ⊕ y]k + [z]k ≤ [(x ⊕ y) ⊕ z]k (x ⊕ y) ⊕ z 2 1 3 ([x] + [y] + ) + ([z] + ) = [x] + [y] + [z] + . k k 10k k 10k k k k 10k 3 Similarly, we can also have [x]k +[y]k +[z]k ≤ [(y ⊕z)⊕x]k ≤ [x]k +[y]k +[z]k + 10k . Finally by Lemma 3.8, (x ⊕ y) ⊕ z ∼ (y ⊕ z) ⊕ x = x ⊕ (y ⊕ z). This concludes the proof. (i) n Remark 3.11. Given n elements {x }i=1 of R and a permutation τ on the index set {1, 2,...,n}, according to Theorems 3.4, 3.9 and 3.10, it is easy to verify that ( (((x(1) ⊕x(2))⊕x(3))⊕x(4)) )⊕x(n) ∼ ( (((x(τ1 ) ⊕x(τ2))⊕x(τ3))⊕x(τ4 )) )⊕x(τn). As usual, we may simply write x(1) ⊕ x(2) ⊕ x(3) ⊕ x(4) ⊕ ⊕ x(n) ∼ x(τ1) ⊕ x(τ2) ⊕ x(τ3) ⊕ x(τ4) ⊕ ⊕ x(τn) since it does not matter where the parentheses lie. 10 LIANGPAN LI 3.2. Additive inverses. Definition 3.12. For any element x = x0.x1x2x3 ∈ R, let Ψ(x) (−1 − x0).(9 − x1)(9 − x2)(9 − x3) , understood as the “additive inverse” of x. The absolute value of x to is defined to be x max{x, Ψ(x)}. Let sign : R → {0, 1} be the signal map 0 if x 0.000000 , sign(x) 1 if x (−1).999999 . Some elementary properties on Ψ( ), and sign( ) are collected below without proofs. The interested readers can easily provide the details without much difficulty. (1) x ⊕ Ψ(x) = (−1).999999 ; (2) Ψ(Ψ(x)) = x; (3) x y ⇔ Ψ(x) Ψ(y); (4) x ∼ y ⇔ Ψ(x) ∼ Ψ(y); (5) x ∼ y ⇒ x ∼ y ; (6) x ∼ Ψ(x) ⇔ x ∼ 0.000000 ; (7) x 0.000000 ; (8) Ψ(x) = x ; (9) sign(x) + sign(Ψ(x)) = 1; (10) Ψ(sign(x))(x)= x ; (11) Ψ(sign(x))( x )= x; (12) Ψ(sup W ) = infΨ(W ); (n) (n) (13) Ψ(LIMIT({(x )}n∈N)) = LIMIT({Ψ(x )}n∈N). Theorem 3.13. Given x,y,z ∈ R with x ⊕ z ∼ y ⊕ z, we have x ∼ y. Proof. By Theorems 3.9 and 3.10, x = x ⊕ 0.000000 ∼ x ⊕ z ⊕ Ψ(z) ∼ y ⊕ z ⊕ Ψ(z) ∼ y ⊕ 0.000000 = y. Theorem 3.14. For any two elements x,y ∈ R, we have Ψ(x ⊕ y) ∼ Ψ(x) ⊕ Ψ(y). Proof. By Theorems 3.4, 3.9 and 3.10, Ψ(x ⊕ y)=Ψ(x ⊕ y) ⊕ 0.000000 ⊕ 0.000000 ∼ Ψ(x ⊕ y) ⊕ (x ⊕ Ψ(x)) ⊕ (y ⊕ Ψ(y)) ∼ Ψ(x ⊕ y) ⊕ (x ⊕ y) ⊕ (Ψ(x) ⊕ Ψ(y)) ∼ Ψ(x) ⊕ Ψ(y). ANEWAPPROACHTOTHEREALNUMBERS 11 4. Multiplicative operations 4.1. Multiplicative operations. Definition 4.1 (Multiplication). For any two elements x,y ∈ R, let sign(x)+sign(y) x ⊗ y Ψ LIMIT {[ x ]k [ y ]k}k∈N , where Ψ(k) is the k-times composites of Ψ. Example 4.2. For any x ∈ R, (sign(x)) (sign(x)) x ⊗ 1.000000 =Ψ LIMIT {[ x ]k}k∈N =Ψ ( x )= x. This means (R, ⊗) has a unit. Example 4.3. For any x ∈ R, sign(x) x ⊗ 0.000000 =Ψ LIMIT {[ x ]k 0}k∈N =Ψ sign(x) (0.000000 ) ∼ 0.000000 , sign(x)+1 x ⊗ (−1).999999 =Ψ LIMIT {[ x ]k 0}k∈N =Ψ sign(x)+1 (0.000000 ) ∼ 0.000000 . Example 4.4. Given a calculator with sufficiently long digits, we could observe that 1 < 1.42 < 1.99 < 2 < 1.52 1.9 < 1.412 < 1.9999 < 2 < 1.422 1.99 < 1.4142 < 1.999999 < 2 < 1.4152 . . 1 1. 99 99 < (a .a a a a )2 < 1. 999 999 < 2 < (a .a a a a + )2 0 1 2 3 n 0 1 2 3 n 10n n−1 2n . | {z } | {z } . For thousands of years a a0.a1a2a3 has been understood as the positive square root of 2, so what is the reason behind? According to Definition 4.1, a⊗a = 1.999999 . Also from the above formulas, it is no hard to observe (see also [14, 23, 25]) that there is no element z ∈ R such that z ⊗ z = 2.000000 . So if we want to define the positive square root of 2, except a0.a1a2a3 , which else could be? Remark 4.5. Given x,y ∈ QF ⊂ Q with x,y 0.000000 , it is easy to verify that x ⊗ y = x y. Therefore from now on we can abuse the uses of ⊗ and if the summands lie in QF with signs zero. Theorem 4.6. For any two elements x,y ∈ R, we have x ⊗ y = y ⊗ x. 12 LIANGPAN LI Proof. sign(x)+sign(y) x ⊗ y =Ψ LIMIT {[ x ]k [ y ]k}k∈N sign(y)+sign(x) =Ψ LIMIT {[ y ]k [ x ]k}k∈N = y ⊗ x. Theorem 4.7. For any two elements x,y ∈ R, we have Ψ(x ⊗ y)=Ψ(x) ⊗ y = x ⊗ Ψ(y). Proof. This is the twin theorem of Theorem 3.14. By Theorem 4.6 it suffices to prove Ψ(x ⊗ y)=Ψ(x) ⊗ y. To this aim we note sign(Ψ(x))+sign(y) Ψ(x) ⊗ y =Ψ LIMIT {[ Ψ(x) ]k [ y ]k}k∈N 1+sign(x)+sign(y) =Ψ LIMIT {[ x ]k [ y ]k}k∈N sign(x)+sign(y) =Ψ Ψ LIMIT {[ x ]k [ y ]k}k∈N ! = Ψ(x ⊗ y), here we have used the fact that Ψ(2) is the identity map. This proves the theorem. Theorem 4.8. Given x, y, z, w ∈ R with 0.000000 x z and 0.000000 y w, we have x ⊗ y z ⊗ w. Proof. This is the twin theorem of Theorem 3.5. ∀k ∈ N we have 0 ≤ [x]k ≤ [z]k and 0 ≤ [y]k ≤ [w]k, which yields [x]k [y]k ≤ [z]k [w]k. By Lemma 2.9, LIMIT {[x]k [y] } LIMIT {[z] [w] } . This proves the theorem. k k∈N k k k∈N Theorem 4.9. Given x, y, z, w ∈ R with x ∼ z and y ∼ w, we have x ⊗ y ∼ z ⊗ w. Proof. Obviously if two equivalent elements x and z have different signs, then we must have {x, z} = {0.000000 , (−1).999999 }. From Example 4.3, x⊗y ∼ 0.000000 ∼ z⊗w. Thus to prove this theorem, we may assume that sign(x) = sign(z), sign(y) = sign(w), which yields sign(x) + sign(y) = sign(z) + sign(w). By Theorem 4.7, Ψ(sign(x)+sign(y))(x ⊗ y)=Ψ(sign(x))(x) ⊗ Ψ(sign(y))(y) = x ⊗ y , Ψ(sign(z)+sign(w))(z ⊗ w)=Ψ(sign(z))(z) ⊗ Ψ(sign(w))(w)= z ⊗ w . Consequently, to prove this theorem we may further assume below x, y, z, w 0.000000 , by which we shall make use of Theorem 4.8. Let M ∈ N be an upper bound for {x, y, z, w}. For any k ∈ N, 1 1 [x ⊗ y] x ⊗ y ([x] + ) ([y] + ) k k 10k k 10k 2 2 5M ≤ ([z] + ) ([w] + ) ≤ [z] [w] + k 10k k 10k k k 10k 5M 5M + 1 (z ⊗ w) ⊕ [z ⊗ w] + , 10k k 10k ANEWAPPROACHTOTHEREALNUMBERS 13 here we have used Lemma 2.14 for x ∼ z and y ∼ w. By symmetry we can also have 5M+1 [z ⊗ w]k ≤ [x ⊗ y]k + 10k . Finally by Lemma 3.8, x ⊗ y ∼ z ⊗ w. This concludes the whole proof. Theorem 4.10. For any three elements x,y,z ∈ R, we have (x ⊗ y) ⊗ z ∼ x ⊗ (y ⊗ z). Proof. By Theorem 4.7, to prove this theorem we may assume that x,y,z 0.000000 , by which we shall also make use of Theorem 4.8. Let M ∈ N be an upper bound for {x,y,z}. For any k ∈ N, 1 1 1 (x ⊗ y) ⊗ z ([x] + ) ([y] + ) ([z] + ) k 10k k 10k k 10k 4M 2 4M 2 ≤ [x] [y] [z] + w ⊕ , k k k 10k 10k where w LIMIT {[x]n [y]n [z]n}n∈N . On the other hand, [x]n [y]n [z]n (x ⊗ y) ⊗ [z]n (x ⊗ y) ⊗ z. Letting n →∞ yields w (x ⊗ y) ⊗ z. With these preparations in hand, for any k ∈ N we 4M 2 4M 2 have [w]k ≤ [(x ⊗ y) ⊗ z]k ≤ [w ⊕ 10k ]k = [w]k + 10k . Similarly, we can also have [w]k ≤ 4M 2 [(y ⊗ z) ⊗ x]k ≤ [w]k + 10k . Finally by Lemma 3.8, (x ⊗ y) ⊗ z ∼ (y ⊗ z) ⊗ x = x ⊗ (y ⊗ z). This concludes the whole proof. Theorem 4.11. For any three elements x,y,z ∈ R, we have (x⊕y)⊗z ∼ (x⊗z)⊕(y ⊗z). Proof. By Theorems 3.14 and 4.7, to prove this theorem we may assume sign(z) = 0, and suppose this is the case. Case 1: Suppose sign(x) = sign(y). By Theorems 3.14 and 4.7 again, we may further assume sign(x) = sign(y) = 0. Let M ∈ N be an upper bound for {x,y,z}. For any k ∈ N, 1 1 [(x ⊕ y) ⊗ z] (x ⊕ y) ⊗ z ([x ⊕ y] + ) ([z] + ) k k 10k k 10k 3 1 6M ≤ ([x] + [y] + ) ([z] + ) ≤ [x] [z] + [y] [z] + k k 10k k 10k k k k k 10k 6M 6M + 1 (x ⊗ z) ⊕ (y ⊗ z) ⊕ [(x ⊗ z) ⊕ (y ⊗ z)] + . 10k k 10k On the other hand, [(x ⊗ z) ⊕ (y ⊗ z)]k (x ⊗ z) ⊕ (y ⊗ z) 1 1 1 1 ([x] + ) ([z] + ) + ([y] + ) ([z] + ) k 10k k 10k k 10k k 10k 6M 6M ≤ ([x] + [y] ) [z] + ((x ⊕ y) ⊗ z) ⊕ k k k 10k 10k 6M + 1 [(x ⊕ y) ⊗ z] + . k 10k Thus by Lemma 3.8, (x ⊕ y) ⊗ z ∼ (x ⊗ z) ⊕ (y ⊗ z). Case 2: Suppose x,y have different signs. Thus x ⊕ y has the same sign as one of x,y, say for example, has the same sign as x. Thus Ψ(y) and x ⊕ y have the same sign. 14 LIANGPAN LI According to the analysis in Case 1, (Ψ(y) ⊕ (x ⊕ y)) ⊗ z ∼ (Ψ(y) ⊗ z) ⊕ ((x ⊕ y) ⊗ z). Adding y ⊗ z to the both sides of the above formula yields (y ⊗ z) ⊕ (x ⊗ z) ∼ (y ⊗ z) ⊕ (Ψ(y) ⊗ z) ⊕ ((x ⊕ y) ⊗ z). So we are left to prove that (y ⊗ z) ⊕ (Ψ(y) ⊗ z) ∼ 0.000000 . To this aim, we note Ψ (y ⊗ z) ⊕ (Ψ(y) ⊗ z) ∼ (Ψ(y) ⊗ z) ⊕ (y ⊗ z) = (y ⊗ z) ⊕ (Ψ(y) ⊗ z), which naturally implies (y ⊗ z) ⊕ (Ψ(y) ⊗ z) ∼ 0.000000 . This concludes the whole proof of the theorem. 4.2. Multiplicative inverses. Definition 4.12 (Inverse of multiplication). For any x ∈ R with x ≻ 0.000000 , let −1 sign(x) 1 x Ψ LIMIT { }k∈N . [ x ]k 1 For some small k, [ x ]k might be equal to zero. So it is possible that is meaningless, [ x ]k or has the meaning of positive infinity. But since we are mainly concerned with the greatest lower bound for the set { 1 | k ∈ N}, it does not matter whether the initial items of the [ x ]k 1 sequence { }k N are equal to positive infinity or not. [ x ]k ∈ Theorem 4.13. For any x ∈ R with x ≻ 0.000000 , we have Ψ(x)−1 = Ψ(x−1). Proof. Since Ψ(x) = x ≻ 0.000000 , by definition we have −1 sign(Ψ(x)) 1 Ψ(x) =Ψ LIMIT { }k∈N [ Ψ(x) ]k 1+sign(x) 1 −1 =Ψ LIMIT { }k∈N = Ψ(x ). [ x ]k Theorem 4.14. For any x ∈ R with x ≻ 0.000000 , we have x⊗x−1 ∼ 1.000000 . Proof. According to Theorem 4.13, to prove this theorem we may assume x ≻ 0.000000 , and suppose this is the case. Since x ≻ 0.000000 , one can find a sufficiently large 1 −n m ∈ N such that 10m x. For any naturals number k ≥ n ≥ m, [x]n ≤ [x]k ≤ [x]n + 10 . 1 1 1 Hence taking reciprocals in each part we get −n ≤ ≤ . Letting k →∞ yields [x]n+10 [x]k [x]n 1 −1 1 −n x . Combining this with Lemma 3.6 we have [x]n+10 [x]n 1 −1 1 1 [x]n 1 x ⊗ x ([x]n + n ) . [x]n + 10n 10 [x]n 1 1 1 m−n Since 10m x, we have 10m ≤ [x]m ≤ [x]n, and consequently, [x]n 1 ≥ 1 − 10 , [x]n+ 10n 1 1 m−n ([x] + n ) ≤ 1 + 10 . Finally for any s ∈ N, let us take n = m + s. Then n 10 [x]n 1 −1 1 1 −1 1 1 − 10s x ⊗ x 1+ 10s , which yields 1 − 10s ≤ [x ⊗ x ]s ≤ 1+ 10s . Note also 1 1 1 − 10s ≤ [1.000000 ]s ≤ 1+ 10s . Thus a standard application of Lemma 2.14 gives the desired result. We are done. ANEWAPPROACHTOTHEREALNUMBERS 15 5. Construction of real numbers 5.1. Motivation. In mathematics, since R is a set, there is no doubt that 0.999999 and 1.000000 are two different elements. But in the real world, if we still imagine 0.999999 and 1.000000 as two different objects, then it will cause some trouble. For example, how can we define and understand 0.999999 + 1.000000 ? 2 Since there is no element z ∈ R such that 0.999999 ≺ z ≺ 1.000000 , we may regard R having “holes” in many places. Have you ever seen a bunch of sunlight having lots of holes? It is too terrible! ∞ Noting the sequence {[0.999999 ]k}k=1 is getting “closer and closer” to 1.000000 as k approaches to infinity, the best way we think to understand the relation between 0.999999 and 1.000000 both in mathematics and in the real world, is to view them as the same object, which leads to the invention of the real “real numbers” below. 5.2. Methodology. Define the set of real numbers R to be the set of equivalent classes R/ ∼. For those readers who are not too familiar with the algebraic representation R/ ∼, the definition of R is not too abstract at all: if the equivalent class has exactly two elements, you may simply discard one element, say for example discard those elements in R9 as we have done in Section 1, and leave the other one remained. For any two equivalent classes [x], [y], we introduce an induced operation ⊕ on R by [x]⊕[y] [x ⊕ y]. b By Theorems 3.9, ⊕ is well-defined. By Theorem 3.4, ⊕ is commutative. By Theorem 3.10, ⊕ is associative. From Example 3.2, (Rb, ⊕) has a unit [0.000000 ]. Since x ⊕ Ψ(x) = (−1).999999 ∼ b0.000000 , every element of (R, ⊕b) has an inverse. In the language of algebra,b (R, ⊕) is an Abelian group. b b For any two equivalent classes [x], [y], we introduce an induced operation ⊗ on R by b [x]⊗[y] [x ⊗ y]. b By Theorem 4.9, ⊗ is well-defined. By Theorem 4.6, ⊗ is commutative. By Theorem 4.10, ⊗ is associative. From Example 4.2, (Rb, ⊗) has a unit [1.000000 ]. By Theorem 4.14, every element of (bR∗, ⊗) has an inverse, where R∗ bR\[0.000000 ]. By Theorem 4.11, ⊗b is distribute over ⊕. Hence we have derivedb a number system (R, ⊕, ⊗), which in the language of algebra isb a field. b We can also introduceb an order on R from the total order onbRb, that is, we say [x] [y] if y ⊕ Ψ(x) (−1).999999 . The interested readers can verify that is well- defined and is a total order. Finally,b we state the least upper and greatest lower bounds propertiesb for (R, ) below without proofs, which is a simple corollary of Theorem 2.5. Theorem 5.1. Every nonempty bounded above (below) subset of (R, ) has a least upper (greatest lower) bound.b b As experienced readers should know, from the least upper bound property, after intro- ducing Weierstrass’s ǫ − N definition of convergence of sequences, we can deduce many other theorems such as the monotone convergence theorem, Cauchy’s convergence princi- ple, the Bolzano-Weierstrass theorem, the Heine-Borel theorem and so on in a few pages. 16 LIANGPAN LI 6. From decimals to Dedekind cuts 6.1. A natural bijection. Definition 6.1. A Dedekind cut is a pair of nonempty subsets A, B of Q, denoted by (A|B), such that • A ∩ B = ∅, A ∪ B = Q, • a ∈ A, b ∈ B ⇒ a < b, • A contains no greatest element. The set of all Dedekind cuts is denoted by DR. Given a Dedekind cut (A|B), we obviously have sup A inf B. To go even further, we can have the following theorem. Theorem 6.2. For any Dedekind cut (A|B), we have sup A ∼ inf B. (1) a+b (1) Proof. Fix arbitrarily k ∈ N, a ∈ A and b ∈ B. Step 1: Let c 2 . If c ∈ A, then we let a(1) c(1), b(1) b; else if c(1) ∈ B, then we let a(1) a, b(1) c(1). Obviously (1) (1) b−a (2) a(1)+b(1) (2) (2) (2) (2) (1) b −a = 2 . Step 2: Let c 2 . If c ∈ A, then we let a c , b b ; (2) (2) (1) (2) (2) (2) (2) b(1)−a(1) b−a else if c ∈ B, then we let a a , b c . Obviously b − a = 2 = 4 . We can repeat this procedure to the 4k-th step to get a(4k) ∈ A and b(4k) ∈ B so that (4k) (4k) b−a (4k) (4k) b − a = 16k . Note a sup A inf B b , thus 1 ⌈b − a⌉ 1 2⌈b − a⌉ 0 ≤ [inf B] −[sup A] ≤ [b(4k)] −[a(4k)] ≤ b(4k)−(a(4k)− ) ≤ + ≤ , k k k k 10k 16k 10k 10k here we have used Lemma 3.6 at the third inequality. By Lemma 3.8, sup A ∼ inf B. Now we can derive a map τ from DR to R/ ∼ by sending (A|B) to [sup A] = [inf B]. It would be great if τ is bijective. In the following we shall prove this is indeed the case. Lemma 6.3. We have R9 ∩ Q = ∅. Proof. Suppose the contrary one can find an element α ∈ R9 ∩ Q. Let us choose the corresponding element β ∈ QF so that α ≺ β, α ∼ β. Obviously as elements of Q, α < β. N 3 Now choose a sufficiently large k ∈ so that β − α ≥ 10k , then by Lemma 3.6, 1 2 [β] − [α] ≥ (β − ) − α ≥ . k k 10k 10k According to Lemma 2.14, this is impossible. We are done. Theorem 6.4. The map τ from DR to R/ ∼ is bijective. Proof. We first prove that τ is surjective. Fix arbitrarily x ∈ R. Let A {q ∈ Q| q ≺ x}, B {q ∈ Q| q x}. Obviously to verify that (A|B) is a Dedekind cut, it suffices to prove that A contains no greatest element. Suppose the contrary, then sup A ∈ A, which by the definitions of A ANEWAPPROACHTOTHEREALNUMBERS 17 and B leads to sup A ≺ x inf B. By Theorem 6.2, sup A ∼ inf B. By Lemma 2.12, sup A ∈ R9, which contradicts to Lemma 6.3. Hence (A|B) must be a Dedekind cut. Note sup A x inf B, by Theorem 6.2 again, we have τ((A|B)) = [x]. Thus τ is surjective. Next we prove that τ is injective. Given two different Dedekind cuts (A|B), (C|D), without loss of generality we may assume there exists an element α ∈ C\A, which by the definition of Dedekind cut yields sup A α ≺ sup C. Our aim is to prove that sup A and sup C are not equivalent. Suppose this is the contrary, then sup A ∼ sup C. By Lemma 2.12, α = sup A ∈ R9, which contradicts to Lemma 6.3. Thus τ is injective. This concludes the whole proof of the theorem. 6.2. Isomorphisms between operations. Definition 6.5. A generalized Dedekind cut is a pair of nonempty subsets A, B of Q, denoted by [A B], such that • A ∩ B = ∅, A ∪ B = Q, • a ∈ A, b ∈ B ⇒ a < b. Remark 6.6. To form a Dedekind cut (A|B) from a generalized Dedekind cut [A B], we can simply move the greatest element of A if it exists, from A to B. In this case it is no hard to prove that sup A ∼ inf B ∼ sup A ∼e infe B, so we can abuse the uses of [A B] and (A|B). For example, let Ψ : DR → DR be the additive inverse map defined by sending (A|B) to [−B −A]. e e Definitione e 6.7. Let Θ : DRe → {0, 1} be the signal map Θ((A|B)) max sign(x). x∈B The absolute value of a Dedekind cut (A|B) is defined by (A|B) Ψ Θ((A|B)) ((A|B)). Definition 6.8. For any two Dedekind cuts (A|B), (C|D), define the addition of both cuts by e (A|B) ⊞ (C|D) = [Q\(B + D) B + D]. If the Dedekind cuts (A|B), (C|D) are of signs zero, define the multiplication of them by (A|B) ⊠ (C|D) = [Q\(B D) B D]. Generally for any two Dedekind cuts (A|B), (C|D), define their multiplication by (A|B) ⊠ (C|D)= Ψ Θ((A|B))+Θ((C|D)) (A|B) ⊠ (C|D) . Theorem 6.9. For any two Dedekinde cuts (A|B), (C|D), we have τ (A|B) ⊞ (C|D) = τ (A|B) ⊕ τ (C|D) . Proof. To prove this theorem, it suffices to prove thatb inf(B + D) ∼ inf B ⊕ inf D. To this aim we note sup(A + C) sup A ⊕ sup C inf B ⊕ inf D inf(B + D). Hence we need only to prove that sup(A + C) ∼ inf(B + D). Fix arbitrarily k ∈ N. We can repeat the procedures in the proof of Theorem 6.2 from any common starting points a ∈ A∩C and b ∈ B ∩D to get a(4k) ∈ A, b(4k) ∈ B, c(4k) ∈ C 18 LIANGPAN LI (4k) (4k) (4k) (4k) (4k) b−a and d ∈ D so that b − a = d − c = 16k . Thus a(4k) + c(4k) sup(A + C) inf(B + D) b(4k) + d(4k), and consequently, 2⌈b − a⌉ 1 0 ≤ [inf(B + D)] − [sup(A + C)] ≤ [b(4k) + d(4k)] − [a(4k) + c(4k)] ≤ + . k k k k 16k 10k By Lemma 3.8, we are done. Theorem 6.10. For any two Dedekind cuts (A|B), (C|D), we have τ (A|B) ⊠ (C|D) = τ (A|B) ⊗ τ (C|D) . Proof. We shall only prove this theorem for the specialb case that the Dedekind cuts (A|B), (C|D) are of signs zero, and leave the general case to the interested readers. Suppose the Dedekind cuts (A|B), (C|D) are of signs zero. To our aim, it suffices to prove that inf(B D) ∼ inf B ⊗ inf D. If either 0 ∈ B or 0 ∈ D, then it is easy to observe that inf(B D) = inf B ⊗ inf D = 0.000000 . Hence we may further assume that 0 ∈ A ∩ C. Let us choose a fixed natural number z ∈ B ∩ D. Fix arbitrarily k ∈ N. We can repeat the procedures in the proof of Theorem 6.2 from the starting points 0 ∈ A ∩ C and z ∈ B ∩ D to get a(4k) ∈ A, b(4k) ∈ B, c(4k) ∈ C and (4k) (4k) (4k) (4k) (4k) z−0 z d ∈ D so that b − a = d − c = 16k = 16k . Thus 0.000000 a(4k) c(4k) sup A ⊗ sup C inf B ⊗ inf D inf(B D) b(4k) d(4k), and consequently, 2z2 z2 1 0 ≤ [inf(B D)] − [inf B ⊗ inf D] ≤ [b(4k) d(4k)] − [a(4k) c(4k)] ≤ + + . k k k k 16k 256k 10k By Lemma 3.8, we are done. Given two Dedekind cuts (A|B), (C|D), we say (A|B) is less than or equal to (C|D), denoted by (A|B) (C|D), if sup A sup C. It is a piece of cake to prove that (A|B) (C|D) ⇔ τ((A|B)) τ((C|D)). Thus with Theorems 6.4, 6.9 and 6.10, (R, ⊕, ⊗, ) is isomorphic to (DR, ⊞, ⊠, ). b 7. From decimals tob b Cauchyb sequences 7.1. A natural surjection. (n) Definition 7.1. A sequence of rational numbers {x }n∈N is called Cauchy if for every rational ǫ> 0, there exists an N ∈ N such that ∀m,n ≥ N, |x(m) − x(n)| < ǫ. The set of all Cauchy sequences of rational numbers is denoted by CR. Two Cauchy sequences of (n) (n) (n) rational numbers {x }n∈N, {y }n∈N are said to be equivalent, denoted by {x }n∈N ≈ (n) (n) {y }n∈N, if for every rational ǫ > 0, there exists an N ∈ N such that ∀n ≥ N, |x − y(n)| <ǫ. It is easy to prove ≈ is an equivalence relation, thus we can get a quotient space CR/ ≈, (n) denoted by CR for consistency. Given a Cauchy sequence of rational numbers {x }n∈N, (n) (n) we obviously have LIMIT({x }n∈N) LIMIT({x }n∈N). To go even further, we can have the followingd theorem. ANEWAPPROACHTOTHEREALNUMBERS 19 (n) (n) Theorem 7.2. LIMIT({x }n∈N) ∼ LIMIT({x }n∈N) holds for any Cauchy sequence (n) of rational numbers {x }n∈N. Proof. Fix arbitrarily k ∈ N, there exists an N ∈ N depending on k such that ∀m,n ≥ N, (m) (n) 1 (N) 1 (n) (N) 1 |x − x | < 10k . Thus ∀n ≥ N, x − 10k ≤ x ≤ x + 10k . Letting n →∞ gives (N) 1 (n) (n) (N) 1 x − LIMIT({x } N) LIMIT({x } N) x + , 10k n∈ n∈ 10k and consequently (n) (n) (N) 1 (N) 1 2 0 ≤ [LIMIT({x } N)] − [LIMIT({x } N)] ≤ [x + ] − [x − ] = . n∈ k n∈ k 10k k 10k k 10k (n) (n) By Lemma 3.8, LIMIT({x }n∈N) ∼ LIMIT({x }n∈N). This finishes the proof. (n) Now we can derive a map κ from CR to R/ ∼ by sending Cauchy sequence {x }n∈N to (n) (n) [LIMIT({x }n∈N)] = [LIMIT({x }n∈N)]. Obviously, κ is surjective since for any x ∈ R, we have a typical Cauchy sequence {[x]k}k∈N such that x = LIMIT {[x]k}k∈N . (n) (n) Theorem 7.3. For any two Cauchy sequences of rational numbers {x }n∈N,{y }n∈N, (n) (n) (n) (n) {x }n∈N ≈ {y }n∈N ⇔ κ({x }n∈N)= κ({y }n∈N). (n) (n) Proof. Suppose {x }n∈N ≈ {y }n∈N. Fix arbitrarily k ∈ N, there exists an N ∈ N such (n) (n) 1 (n) (n) 1 that ∀n ≥ N, |x − y | < 10k . Equivalently, ∀n ≥ N we have x ≤ y + 10k and (n) (n) 1 (n) (n) 1 y ≤ x + 10k . Letting n → ∞ yields LIMIT({x }n∈N) LIMIT({y }n∈N) ⊕ 10k (n) (n) 1 and LIMIT({y }n∈N) LIMIT({x }n∈N) ⊕ 10k , which further implies (n) (n) 1 [LIMIT({x } N)] − [LIMIT({y } N)] ≤ . n∈ k n∈ k 10k (n) (n) By Lemma 3.8, LIMIT( {x }n∈N) ∼ LIMIT({y }n∈N). This proves the necessary part. (n) (n) Suppose κ({x }n∈N) = κ({y }n∈N). We argue by contradiction the sufficient part (n) (n) and suppose {x }n∈N, {y }n∈N are not equivalent. Then there exist a positive rational number ǫ0 and a sequence of natural numbers n1 < n2 < n3 < such that ∀i ∈ N, (ni) (ni) (ni) (ni) (ni) (ni) |x − y |≥ ǫ0. It may happen either x ≥ y + ǫ0 or y ≥ x + ǫ0. Without (ni) (ni) loss of generality, we may assume there are infinitely many terms x ≥ y + ǫ0. Now N 2 we first choose a k ∈ such that 10k ≤ ǫ0, then choose a subsequence {nij }j∈N from (nij ) (nij ) 2 {ni}i∈N such that x ≥ y + 10k . Thus (ni ) (ni ) 2 LIMIT({x j }j N) LIMIT({y j }j N) ⊕ , ∈ ∈ 10k (nij ) (nij ) 2 which yields [LIMIT({x }j∈N)]k ≥ [LIMIT({y }j∈N)]k + 10k , and consequently by (ni ) (ni ) Lemma 3.8, LIMIT({x j }j∈N) and LIMIT({y j }j∈N) are not equivalent in R. But this is impossible since by Lemma 2.9, Theorem 7.2 and our assumptions, (n) (ni ) (ni ) (n) LIMIT({x }n∈N) LIMIT({x j }j∈N) LIMIT({x j }j∈N) LIMIT({x }n∈N), (n) (ni ) (ni ) (n) LIMIT({y }n∈N) LIMIT({y j }j∈N) LIMIT({y j }j∈N) LIMIT({y }n∈N), (n) (n) (n) (n) LIMIT({x }n∈N) ∼ LIMIT({x }n∈N) ∼ LIMIT({y }n∈N) ∼ LIMIT({y }n∈N), 20 LIANGPAN LI (ni ) (ni ) which naturally implies that LIMIT({x j }j∈N) ∼ LIMIT({y j }j∈N). This proves the sufficient part, also concludes the whole proof of the theorem. 7.2. Homomorphisms between operations. (n) (n) Definition 7.4. Given two Cauchy sequences of rational numbers {x }n∈N, {y }n∈N, (n) (n) (n) (n) it is easy to verify that {x + y }n∈N and {x y }n∈N are also Cauchy sequences, so we can define the addition and multiplication respectively by (n) (n) (n) (n) {x }n∈N⊞ {y }n∈N = {x + y }n∈N, (n) (n) (n) (n) {x }n∈N⊠ {y }n∈N = {x y }n∈N. b (n) (n) Theorem 7.5. For any two Cauchyb sequences of rational numbers {x }n∈N, {y }n∈N, (n) (n) (n) (n) κ {x }n∈N⊞ {y }n∈N = κ {x }n∈N ⊕ κ {y }n∈N . Proof. Fix arbitrarily k ∈ Nb, there exists a common startingb index N ∈ N such that (m) (n) 1 (m) (n) 1 ∀m,n ≥ N, |x − x | < 10k , |y − y | < 10k . Thus ∀n ≥ N, 1 1 x(N) − ≤ x(n) ≤ x(N) + , 10k 10k 1 1 y(N) − ≤ y(n) ≤ y(N) + , 10k 10k 2 2 x(N) + y(N) − ≤ x(n) + y(n) ≤ x(N) + y(N) + . 10k 10k Letting n →∞ gives (N) 1 (n) (N) 1 (7.1) x − LIMIT({x } N) x + , 10k n∈ 10k (N) 1 (n) (N) 1 (7.2) y − LIMIT({y } N) y + , 10k n∈ 10k (N) (N) 2 (n) (n) (N) (N) 2 (7.3) x + y − LIMIT({x + y } N) x + y + . 10k n∈ 10k Applying Theorem 3.5 to (7.1) and (7.2) yields (7.4) (N) (N) 2 (n) (n) (N) (N) 2 x + y − LIMIT({x }n N) ⊕ LIMIT({y }n N) x + y + . 10k ∈ ∈ 10k Finally a standard application of Lemma 3.8 to (7.3) and (7.4) gives the desired result, we are done. (n) (n) Theorem 7.6. For any two Cauchy sequences of rational numbers {x }n∈N, {y }n∈N, (n) (n) (n) (n) κ {x }n∈N⊠ {y }n∈N = κ {x }n∈N ⊗ κ {y }n∈N . Proof. By Formula (13) in Subsectionb 3.2 and Theorem 4.7,b to prove this theorem we may (n) (n) assume both LIMIT({x }n∈N) 0.000000 and LIMIT({y }n∈N) 0.000000 , (n) (n) which easily implies that {x }n∈N and {y }n∈N are eventually positive, that is, there is an starting index M ∈ N such that ∀n ≥ M, x(n) 0.000000 , y(n) 0.000000 . ANEWAPPROACHTOTHEREALNUMBERS 21 Similar to the proof of the previous theorem, for arbitrary k ∈ N one can find a starting (m) (n) 1 (m) (n) 1 index N ≥ M such that ∀m,n ≥ N, |x − x | < 10k , |y − y | < 10k , and also (N) 1 (n) (N) 1 (7.5) x − LIMIT({x } N) x + , 10k n∈ 10k (N) 1 (n) (N) 1 (7.6) y − LIMIT({y } N) y + , 10k n∈ 10k (N) (N) 2z 1 (n) (n) (N) (N) 2z 1 (7.7) x y − − LIMIT({x y }n N) x y + + , 10k 100k ∈ 10k 100k (n) (n) where z ∈ N is any common upper bound for the sequences {|x |}n∈N, {|y |}n∈N. (N) 1 (N) 1 Case 1: Suppose x − 10k ≥ 0 and y − 10k ≥ 0. Then applying Theorem 4.8 to (7.5) and (7.6) yields 2z 1 (n) (n) 2z 1 (7.8) s − − LIMIT({x } N) ⊗ LIMIT({y } N) s + + , 10k 100k n∈ n∈ 10k 100k where s stands for x(N) y(N) purely for the sake of simplicity. Next a standard application of Lemma 3.8 to (7.7) and (7.8) gives the desired result. (N) 1 Case 2: Suppose x − 10k < 0. Now we revise (7.5)∼(7.7) slightly to (n) 2 (7.9) 0 LIMIT({x } N) , n∈ 10k (n) 1 (7.10) 0 LIMIT({y } N) z + , n∈ 10k (n) (n) 3z 1 (7.11) 0 LIMIT({x y } N) + . n∈ 10k 100k Then applying Theorem 4.8 to (7.9) and (7.10) yields (n) (n) 2z 2 (7.12) 0 LIMIT({x }n N) ⊗ LIMIT({y }n N) + , ∈ ∈ 10k 100k Finally a standard application of Lemma 3.8 to (7.11) and (7.12) gives the desired result. (N) 1 Case 3: Suppose y − 10k < 0. The proof is fully identical to that in Case 2. This concludes the whole proof of the theorem. (n) (n) (n) Given two Cauchy sequences of rational numbers {x }n∈N, {y }n∈N, we say {x }n∈N (n) (n) (n) (n) is less than or equal to {y }n∈N, denoted by {x }n∈N {y }n∈N, if either {x }n∈N ≈ (n) (n) (n) {y }n∈N or the sequence {y − x }n∈N is eventually positive, that is, there exists an n ∈ N such that ∀n ≥ N, y(n) − x(n) > 0. It is also veryb easy to verify that (n) (n) (n) (n) {x }n∈N {y }n∈N ⇔ κ({x }n∈N) κ({y }n∈N). Thus with Theorem 3.9, Theorem 4.9, Theorem 7.3, Theorem 7.5 and Theorem 7.6, (R, ⊕, ⊗, ) is isomorphic tob (CR, ⊞, ⊠, ). b We note a noteworthy difference. When constructing the real numbers via decimals as before,b b web introduced a total orderd asb earlierb b as possible, then based on derived operations of this order, we introduced additive and multiplicative operations. But when people work on the Cauchy sequence approach, total order is not so important a concept as we regard. What they prefer most is another fundamental concept called “distance”. It is amazing to see different approaches lead to the same structure, mathematics is so harmonious! 22 LIANGPAN LI 8. Rationals v.s. Irrationals According to Lemma 6.3, there are generally three types of elements in R, that is, • Type 1: Q • Type 2: R9 • Type 3: The others remained According to Lemma 2.12, every element of second type will be identified with an element of first type in R. Thus no matter in R or in R, we should pay attention to the third type of elements, which has deserved not so better a name in history. Definition 8.1. An element of R is called irrational if it does not belong to the first two types of elements. Theorem 8.2. An element x0.x1x2x3 ∈ R is irrational if and only if it cannot end with infinite recurrence of a block of digits. q Proof. By the pigeonhole principle, it is easy to prove that any rational number α = p with p ∈ N and q ∈ Z must end with infinite recurrence of a block of length ≤ p. Obviously, any element of R9 is also of such property. Thus to prove this theorem, it suffices to show any element ends with infinite recurrence of a block of digits must belong to the first two types. To this aim, without loss of generality let us suppose x = x0.x1 xk y1 ys y1 ys y1 ys y1 ys y1 ys y1 ys ∈ R with ys < 9 (please think why we can impose this condition), then we need only prove x belongs to the first type,| that{z is,} |x ∈{zQ.} | Now{z we} | define{z } a| rational{z } | number{z } s k s s−j k−i s s−j yj 10 ( xi 10 ) (10 − 1) + yj 10 1 j=1 i=0 j=1 z [x]k + X s = X X . 10k 10 − 1 10k (10s − 1) Then in the following we devise two sequences of rational numbers with left hand { n}n∈N, 1 and right hand {ωn}n∈N satisfying the desired inequalities x0.x1 xk y1 ys ≤ z ≤ x0.x1 xk y1 ys−1(ys + 1) z }| { x .x x y y y y ≤ z ≤ x .x x y y y y (y + 1) 0 1 k 1 s |1 {z }s 0 1 k 1 s 1 s−1 s z }| { x .x x y y y y y y ≤ z ≤ x .x x y y y y y y (y + 1) 0 1 k 1 s |1 {z }s |1 {z }s 0 1 k |1 {z }s 1 s 1 s−1 s ≤ z ≤ . z }| { Note { n}n∈|N is{z monotonically} | {z } | {z increasing} with limit x|, while{z } |{ω{zn}n}∈N is monotonically decreasing with limit also x, thus by Lemma 2.9 we must have x = z. This proves x is a rational number, we are done. 1Let us see from a special example to know why these inequalities are true, from which one can easily 23 prove the general cases. For example, suppose x = 0.232323 and z = 99 . Take the third of this sequence 99 · · · of inequalities for example, multiplying each parts by 23 gives 99 0.999999 1 0.999999 + 0.000001 , ≤ ≤ · 23 which is obviously true. ANEWAPPROACHTOTHEREALNUMBERS 23 9. Algebraic abstraction: Complete ordered field After detailed studies of three explicit approaches to the real number system, let us explain how can we come to its algebraic characterization by the so-called concept of complete ordered field. If there exists only one real number system in the mathematical world, then it should be a system (R, +, ×, ≤) with three binary operations +, ×, ≤ endowed on the same ambient set R satisfying at least a few conditions: • R is infinite (reason: should contain Q as a subset) • (R, +, ×) is a field (reason: (R, ⊕, ⊗) is a field) • ≤ is a total order on R and (R, ≤) has both the least upper bound property and the greatest lower bound property (reason:b b (R, ) is of such properties) An algebraic system satisfies the above conditions can not characterize the real number system yet. The main reason is we can not senseb any difference between the positive and the negative parts of our desired real number system, so we need more observation. To mention a few: addition of elements of CR preserve order , so are multiplication by positive elements of CR. d b Is an algebraic system (R, +, ×, ≤) satisfies the following conditions isomorphic to those systems of Dedekindd and M´eray-Cantor-Heine? • (R, +, ×) is an infinite field • ≤ is a total order on R and (R, ≤) has both the least upper bound property and the greatest lower bound property • x 10. Epilog: Analysis v.s. Algebra or Analysis plus Algebra ? When to teach mathematical majors Mathematical Analysis, instructors need to make a serious choice. Dedekind cuts “Although the definition is geometrically motivated, and is based on a ‘natural idea’, the definition is actually far too abstract to be efficient at the beginning of a calculus course.” ([38]) We also don’t prefer this approach. The main reason is, once the real number system is established through this approach, we can find almost no use of the language of Dedekind cuts in the subsequent study of the Mathematical Analysis course. Cauchy sequences 24 LIANGPAN LI We believe this approach is too abstract to be acceptable for beginners, who have not received adequate mathematical training yet. We also have a sense that when to construct the real number system explicitly, people would prefer more the Dedekind cut approach than the abstract Cauchy sequence one. Introducing it in the Functional Analysis course would be a better choice. Axiomatic definition As said in the Introduction, we found many authors liked to choose, say for example the algebraic-axiomatic definition, then proved that every real number has a suitable decimal representation. But how can we persuade students believe that a real number is an element of an algebraic system satisfying a dozen or so properties at the beginning of a Mathematical Analysis course? Suggestion: Decimal (First) plus Axiomatic (Second) approaches This paper provides a complete approach to the real number system via rather old decimal representations, which perfectly matches what we have learnt a number is in high school. Once the real number system is established through this approach, the least upper bound property, the greatest lower bound property, the supremum, infimum, upper limit and lower limit operations are immediately, also naturally obtained. This shows huge advantages over the Dedekind cut and the Cauchy sequence approaches. 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